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Ch. 3 Solutions

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Basic

B-1. Euler-Lagrange Equation from the Lagrangian Density (Klein-Gordon Field)

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Problem:

The Lagrangian density for a real scalar field \(\phi(x)\) is given by

\[ \mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi - \frac{1}{2}m^2\phi^2 \]

Using the Euler-Lagrange equation (3.7), explicitly compute each of the following steps.

(a) Find \(\dfrac{\partial \mathcal{L}}{\partial \phi}\).

(b) Find \(\dfrac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\).

(c) Substitute the results of (a) and (b) into the Euler-Lagrange equation to derive the Klein-Gordon equation.

(a) Calculation of \(\dfrac{\partial \mathcal{L}}{\partial \phi}\):

In \(\mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi - \frac{1}{2}m^2\phi^2\), the only term containing \(\phi\) itself (without derivatives) is \(-\frac{1}{2}m^2\phi^2\).

\[ \boxed{\frac{\partial \mathcal{L}}{\partial \phi} = -m^2\phi} \]

(b) Calculation of \(\dfrac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\):

Rewriting the kinetic term as \(\frac{1}{2}\eta^{\alpha\beta}\partial_\alpha\phi\,\partial_\beta\phi\) and differentiating with respect to \(\partial_\mu\phi\):

\[ \frac{\partial}{\partial(\partial_\mu\phi)}\left[\frac{1}{2}\eta^{\alpha\beta}\partial_\alpha\phi\,\partial_\beta\phi\right] = \frac{1}{2}\eta^{\alpha\beta}(\delta^\mu_\alpha\,\partial_\beta\phi + \partial_\alpha\phi\,\delta^\mu_\beta) = \frac{1}{2}(\eta^{\mu\beta}\partial_\beta\phi + \eta^{\alpha\mu}\partial_\alpha\phi) = \partial^\mu\phi \]
\[ \boxed{\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} = \partial^\mu\phi} \]

(c) Derivation of the Klein-Gordon equation:

Substituting into the Euler-Lagrange equation \(\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right) - \frac{\partial\mathcal{L}}{\partial\phi} = 0\):

\[ \partial_\mu(\partial^\mu\phi) - (-m^2\phi) = 0 \]
\[ \boxed{\partial_\mu\partial^\mu\phi + m^2\phi = 0} \]

This is the Klein-Gordon equation.

Verification: Substituting a plane wave \(\phi \propto e^{-ik\cdot x}\) gives \(-k_\mu k^\mu + m^2 = 0\), i.e., \(E^2 = |\mathbf{p}|^2 + m^2\), which is consistent with the relativistic dispersion relation.

B-2. Expanding the Indices of the Kinetic Term

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Problem:

Explicitly expand the kinetic term \(\frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi\) of the Lagrangian density into temporal and spatial components using the Minkowski metric \(\eta^{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1)\), and show that

\[ \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi = \frac{1}{2}\dot{\phi}^2 - \frac{1}{2}(\nabla\phi)^2 \]

where \(\dot{\phi} = \partial_0\phi\) and \(\nabla\phi = (\partial_1\phi, \partial_2\phi, \partial_3\phi)\).

Using \(\partial^\mu\phi = \eta^{\mu\nu}\partial_\nu\phi\), we verify each component:

\[ \partial^0\phi = \eta^{00}\partial_0\phi = (+1)\partial_0\phi = \dot{\phi} \]
\[ \partial^i\phi = \eta^{ii}\partial_i\phi = (-1)\partial_i\phi = -\partial_i\phi \quad (i = 1, 2, 3) \]

Summing over \(\mu\):

\[ \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi = \frac{1}{2}\left[\partial_0\phi\,\partial^0\phi + \sum_{i=1}^{3}\partial_i\phi\,\partial^i\phi\right] \]
\[ = \frac{1}{2}\left[\dot{\phi}\cdot\dot{\phi} + \sum_{i=1}^{3}\partial_i\phi\cdot(-\partial_i\phi)\right] \]
\[ \boxed{\frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi = \frac{1}{2}\dot{\phi}^2 - \frac{1}{2}(\nabla\phi)^2} \]

Verification: From the metric signature \((+,-,-,-)\), we can confirm that the temporal component carries a positive sign and the spatial components carry negative signs.

B-3. Dispersion Relation for a Massless Field

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Problem:

From the Lagrangian density \(\mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi\), substitute the plane wave solution

\[ \phi(x) = A\,e^{-ik_\mu x^\mu} = A\,e^{-iEt + i\mathbf{p}\cdot\mathbf{x}} \]

into the resulting equation of motion, and derive the dispersion relation \(E^2 = |\mathbf{p}|^2\). Here, \(k^\mu = (E, \mathbf{p})\).

The equation of motion obtained from the Lagrangian density \(\mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi\) with \(m = 0\) is:

\[ \partial_\mu\partial^\mu\phi = 0 \quad (\text{wave equation}) \]

Computing each derivative of the plane wave solution \(\phi(x) = A\,e^{-iEt + i\mathbf{p}\cdot\mathbf{x}}\):

\[ \partial_0\phi = -iE\,\phi, \qquad \partial_0^2\phi = (-iE)^2\phi = -E^2\phi \]
\[ \partial_i\phi = ip_i\,\phi, \qquad \nabla^2\phi = (ip_1)^2\phi + (ip_2)^2\phi + (ip_3)^2\phi = -|\mathbf{p}|^2\phi \]

Computing the d'Alembertian:

\[ \partial_\mu\partial^\mu\phi = \partial_0^2\phi - \nabla^2\phi = -E^2\phi - (-|\mathbf{p}|^2\phi) = (-E^2 + |\mathbf{p}|^2)\phi = 0 \]

Since \(\phi \neq 0\):

\[ \boxed{E^2 = |\mathbf{p}|^2} \]

Verification: This corresponds to the dispersion relation \(E = |\mathbf{p}|\) for massless particles (such as photons), which is physically correct.

B-4. Lagrangian with Interaction Term

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Problem:

Given that the Lagrangian density of a real scalar field is

\[ \mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi - \frac{1}{2}m^2\phi^2 - \frac{\lambda}{4!}\phi^4 \]

(where \(\lambda\) is the coupling constant), derive the Euler-Lagrange equation and show that

\[ (\partial_\mu\partial^\mu + m^2)\phi + \frac{\lambda}{3!}\phi^3 = 0 \]
\[ \mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi - \frac{1}{2}m^2\phi^2 - \frac{\lambda}{4!}\phi^4 \]

Step 1: Partial derivative with respect to \(\phi\)

\[ \frac{\partial\mathcal{L}}{\partial\phi} = -m^2\phi - \frac{\lambda}{4!}\cdot 4\phi^3 = -m^2\phi - \frac{\lambda}{3!}\phi^3 \]

Step 2: Partial derivative with respect to \(\partial_\mu\phi\)

The \(\phi^4\) term does not contain \(\partial_\mu\phi\), so the result is the same as D1(b):

\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)} = \partial^\mu\phi \]

Step 3: Substitution into the Euler-Lagrange equation

\[ \partial_\mu(\partial^\mu\phi) - \left(-m^2\phi - \frac{\lambda}{3!}\phi^3\right) = 0 \]
\[ \boxed{(\partial_\mu\partial^\mu + m^2)\phi + \frac{\lambda}{3!}\phi^3 = 0} \]

Verification: Setting \(\lambda = 0\) recovers the free Klein-Gordon equation. Also, differentiating \(\frac{\lambda}{4!}\phi^4\) with respect to \(\phi\) gives \(\frac{4\lambda}{4!}\phi^3 = \frac{\lambda}{3!}\phi^3\), confirming that the factorial manipulation is correct.

B-5. Component Expression of the d'Alembertian

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Problem:

Write out the d'Alembertian \(\Box \equiv \partial_\mu\partial^\mu\) explicitly in terms of time and spatial derivatives. Furthermore, write the Klein-Gordon equation \((\Box + m^2)\phi = 0\) in \((t, x, y, z)\) components.

Expanding \(\Box \equiv \partial_\mu\partial^\mu = \eta^{\mu\nu}\partial_\mu\partial_\nu\):

\[ \Box = \eta^{00}\partial_0\partial_0 + \eta^{11}\partial_1\partial_1 + \eta^{22}\partial_2\partial_2 + \eta^{33}\partial_3\partial_3 \]
\[ = (+1)\frac{\partial^2}{\partial t^2} + (-1)\frac{\partial^2}{\partial x^2} + (-1)\frac{\partial^2}{\partial y^2} + (-1)\frac{\partial^2}{\partial z^2} \]
\[ \boxed{\Box = \frac{\partial^2}{\partial t^2} - \nabla^2 = \frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial y^2} - \frac{\partial^2}{\partial z^2}} \]

Writing the Klein-Gordon equation \((\Box + m^2)\phi = 0\) in components:

\[ \boxed{\frac{\partial^2\phi}{\partial t^2} - \frac{\partial^2\phi}{\partial x^2} - \frac{\partial^2\phi}{\partial y^2} - \frac{\partial^2\phi}{\partial z^2} + m^2\phi = 0} \]

Verification: Setting \(m = 0\) reduces this to the ordinary wave equation \(\frac{\partial^2\phi}{\partial t^2} = \nabla^2\phi\).

B-6. Component Expansion of the Continuity Equation

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Problem:

Suppose a conserved current \(j^\mu = (j^0, j^1, j^2, j^3)\) satisfies \(\partial_\mu j^\mu = 0\).

(a) Separate this equation into temporal and spatial components and write it in the form of the continuity equation.

(b) Using Gauss's theorem, show that the conserved charge \(Q = \int d^3x\,j^0\) is time-independent. Assume that \(\mathbf{j}\) falls off sufficiently rapidly at spatial infinity.

(a) Continuity Equation:

Expanding \(\partial_\mu j^\mu = 0\):

\[ \partial_\mu j^\mu = \partial_0 j^0 + \partial_1 j^1 + \partial_2 j^2 + \partial_3 j^3 = \frac{\partial j^0}{\partial t} + \nabla\cdot\mathbf{j} = 0 \]
\[ \boxed{\frac{\partial j^0}{\partial t} + \nabla\cdot\mathbf{j} = 0} \]

This has the same form as the continuity equation for charge density \(\rho = j^0\) and current density \(\mathbf{j}\).

(b) Time Independence of the Conserved Charge:

\[ \frac{dQ}{dt} = \frac{d}{dt}\int d^3x\,j^0 = \int d^3x\,\frac{\partial j^0}{\partial t} \]

Substituting the result from (a), \(\frac{\partial j^0}{\partial t} = -\nabla\cdot\mathbf{j}\):

\[ \frac{dQ}{dt} = -\int d^3x\,\nabla\cdot\mathbf{j} \]

Applying Gauss's theorem:

\[ \frac{dQ}{dt} = -\oint_{\partial V}\mathbf{j}\cdot d\mathbf{S} \]

Assuming that \(\mathbf{j}\) falls off sufficiently rapidly at spatial infinity (\(|\mathbf{j}| \to 0\) as \(|\mathbf{x}| \to \infty\)), the surface integral vanishes when the integration region is taken over all space:

\[ \boxed{\frac{dQ}{dt} = 0} \]

Therefore, \(Q\) is a conserved quantity that does not depend on time.

Consistency check: As a dimensional analysis check, if \(j^0\) has the dimensions of charge density, then \(Q = \int d^3x\,j^0\) has the dimensions of charge, which is consistent.

B-7. Partial Derivatives of the Complex Scalar Field

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Problem:

We treat the complex scalar field \(\phi(x)\) and its complex conjugate \(\phi^*(x)\) as independent variables. Given the Lagrangian density

\[ \mathcal{L} = \partial_\mu\phi^*\,\partial^\mu\phi - m^2\phi^*\phi \]

compute the following.

(a) \(\dfrac{\partial \mathcal{L}}{\partial \phi^*}\)

(b) \(\dfrac{\partial \mathcal{L}}{\partial(\partial_\mu \phi^*)}\)

(c) Write down the Euler-Lagrange equation for \(\phi^*\).

\[ \mathcal{L} = \partial_\mu\phi^*\,\partial^\mu\phi - m^2\phi^*\phi = \eta^{\alpha\beta}\partial_\alpha\phi^*\,\partial_\beta\phi - m^2\phi^*\phi \]

(a) Calculation of \(\dfrac{\partial \mathcal{L}}{\partial \phi^*}\):

The only term containing \(\phi^*\) itself (without derivatives) is \(-m^2\phi^*\phi\). We treat \(\phi\) as a constant:

\[ \boxed{\frac{\partial\mathcal{L}}{\partial\phi^*} = -m^2\phi} \]

(b) Calculation of \(\dfrac{\partial \mathcal{L}}{\partial(\partial_\mu \phi^*)}\):

We differentiate \(\mathcal{L} = \eta^{\alpha\beta}\partial_\alpha\phi^*\,\partial_\beta\phi - m^2\phi^*\phi\) with respect to \(\partial_\mu\phi^*\). Since \(\partial_\beta\phi\) does not depend on \(\partial_\mu\phi^*\), it can be treated as a constant:

\[ \frac{\partial}{\partial(\partial_\mu\phi^*)}\left[\eta^{\alpha\beta}\partial_\alpha\phi^*\,\partial_\beta\phi\right] = \eta^{\alpha\beta}\delta^\mu_\alpha\,\partial_\beta\phi = \eta^{\mu\beta}\partial_\beta\phi = \partial^\mu\phi \]
\[ \boxed{\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi^*)} = \partial^\mu\phi} \]

(c) Euler-Lagrange equation for \(\phi^*\):

\[ \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi^*)}\right) - \frac{\partial\mathcal{L}}{\partial\phi^*} = 0 \]
\[ \partial_\mu(\partial^\mu\phi) - (-m^2\phi) = 0 \]
\[ \boxed{(\partial_\mu\partial^\mu + m^2)\phi = 0} \]

The Klein-Gordon equation for \(\phi\) is obtained.

Verification: Similarly, deriving the Euler-Lagrange equation for \(\phi\) yields \((\partial_\mu\partial^\mu + m^2)\phi^* = 0\), confirming that \(\phi\) and \(\phi^*\) each independently satisfy the Klein-Gordon equation.

B-8. Construction of Noether Current (Application of the Formula)

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Problem:

According to Noether's theorem, when the Lagrangian density is invariant (\(\delta\mathcal{L} = 0\)) under an infinitesimal transformation \(\phi \to \phi + \delta\phi\) of the field \(\phi\), the conserved current is given by

\[ j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta\phi \]

For the Klein-Gordon field \(\mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi - \frac{1}{2}m^2\phi^2\), consider an infinitesimal constant shift \(\delta\phi = \epsilon\) (where \(\epsilon\) is an infinitesimal constant).

(a) Compute \(\delta\mathcal{L}\) under this transformation and verify that this is not a symmetry when \(m \neq 0\).

(b) Verify that this transformation is a symmetry when \(m = 0\), and write down the corresponding conserved current \(j^\mu\).

(a) Calculation of \(\delta\mathcal{L}\):

Under a constant infinitesimal shift \(\delta\phi = \epsilon\), we have \(\partial_\mu(\delta\phi) = 0\), so \(\delta(\partial_\mu\phi) = 0\).

\[ \delta\mathcal{L} = \frac{\partial\mathcal{L}}{\partial\phi}\,\delta\phi + \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta(\partial_\mu\phi) = (-m^2\phi)\cdot\epsilon + \partial^\mu\phi\cdot 0 = -m^2\phi\,\epsilon \]

Since \(\delta\mathcal{L} \neq 0\) when \(m \neq 0\) and \(\phi \neq 0\):

\[ \boxed{\text{For } m \neq 0\text{, the constant shift is not a symmetry}} \]

(b) The case \(m = 0\):

When \(m = 0\):

\[ \delta\mathcal{L} = -0\cdot\phi\,\epsilon = 0 \]

Therefore the constant shift is a symmetry. The corresponding conserved current is:

\[ j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta\phi = \partial^\mu\phi\cdot\epsilon \]

Factoring out \(\epsilon\) (since \(\epsilon\) is an arbitrary infinitesimal constant, we extract the coefficient):

\[ \boxed{j^\mu = \partial^\mu\phi} \]

Verification: The equation of motion for \(m = 0\) is \(\partial_\mu\partial^\mu\phi = 0\), so \(\partial_\mu j^\mu = \partial_\mu\partial^\mu\phi = 0\) is indeed satisfied. For \(m \neq 0\), we have \(\partial_\mu j^\mu = \partial_\mu\partial^\mu\phi = -m^2\phi \neq 0\), which is consistent with the current not being conserved.

Medium

M-1. Derivation of Conjugate Momentum and Hamiltonian Density

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Problem:

For the Lagrangian density of a real scalar field \(\mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi - \frac{1}{2}m^2\phi^2\), perform the following.

(a) Define the momentum density \(\pi(x)\) conjugate to the field \(\phi\) by

\[ \pi(x) \equiv \frac{\partial\mathcal{L}}{\partial\dot{\phi}} \]

and find its explicit expression.

(b) Construct the Hamiltonian density \(\mathcal{H}\) via the Legendre transformation

\[ \mathcal{H} = \pi\dot{\phi} - \mathcal{L} \]

and write it in terms of \(\pi\), \(\phi\), and \(\nabla\phi\).

(c) Explain the physical meaning of each term in the resulting \(\mathcal{H}\) by analogy with the particle mechanics Hamiltonian \(H = T + V\).

(a) Conjugate Momentum Density:

\(\mathcal{L} = \frac{1}{2}\dot{\phi}^2 - \frac{1}{2}(\nabla\phi)^2 - \frac{1}{2}m^2\phi^2\) (using the result from D2)

\[ \pi(x) \equiv \frac{\partial\mathcal{L}}{\partial\dot{\phi}} = \frac{\partial}{\partial\dot{\phi}}\left[\frac{1}{2}\dot{\phi}^2\right] = \dot{\phi} \]
\[ \boxed{\pi(x) = \dot{\phi}(x)} \]

(b) Hamiltonian Density:

We compute the Legendre transformation \(\mathcal{H} = \pi\dot{\phi} - \mathcal{L}\). Using \(\dot{\phi} = \pi\):

\[ \mathcal{H} = \pi\cdot\pi - \left[\frac{1}{2}\pi^2 - \frac{1}{2}(\nabla\phi)^2 - \frac{1}{2}m^2\phi^2\right] \]
\[ = \pi^2 - \frac{1}{2}\pi^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{1}{2}m^2\phi^2 \]
\[ \boxed{\mathcal{H} = \frac{1}{2}\pi^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{1}{2}m^2\phi^2} \]

(c) Physical Interpretation:

By analogy with the particle mechanics Hamiltonian \(H = T + V\) (kinetic energy + potential energy):

Term Physical meaning Correspondence to particle mechanics
\(\frac{1}{2}\pi^2 = \frac{1}{2}\dot{\phi}^2\) Energy density associated with the time variation of the field Kinetic energy \(\frac{1}{2}m\dot{q}^2\)
\(\frac{1}{2}(\nabla\phi)^2\) Energy density associated with spatial gradients of the field (the cost of neighboring points having different field values) Spring potential energy in a system of masses connected by springs
\(\frac{1}{2}m^2\phi^2\) Energy density associated with the field value itself (mass term) Harmonic oscillator potential \(\frac{1}{2}k q^2\)

An important point is that all three terms in \(\mathcal{H}\) are positive definite, so the energy density satisfies \(\mathcal{H} \geq 0\). This guarantees the stability of the theory.

Consistency check: We verify the dimensions of \(\mathcal{H}\). In natural units, \([\phi] = \text{mass}^1\) and \([\partial_\mu] = \text{mass}^1\), so \([\mathcal{H}] = \text{mass}^4\), which matches the dimensions of energy density (= energy/volume = mass\(^4\)).

M-2. Internal Symmetry and Noether Current of the Complex Scalar Field

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Problem:

The Lagrangian density of the complex scalar field

\[ \mathcal{L} = \partial_\mu\phi^*\,\partial^\mu\phi - m^2\phi^*\phi \]

is invariant under the global \(U(1)\) transformation

\[ \phi(x) \to e^{i\alpha}\phi(x), \qquad \phi^*(x) \to e^{-i\alpha}\phi^*(x) \]

where \(\alpha\) is a real constant.

(a) Using the infinitesimal transformation (\(\alpha \ll 1\)) with \(\delta\phi = i\alpha\phi\) and \(\delta\phi^* = -i\alpha\phi^*\), explicitly verify that \(\delta\mathcal{L} = 0\).

(b) Using Noether's theorem, derive the conserved current \(j^\mu\).

(c) Write down the conserved charge \(Q = \int d^3x\,j^0\) and qualitatively explain why it corresponds to "particle number \(-\) antiparticle number."

(a) Verification that \(\delta\mathcal{L} = 0\):

Under the infinitesimal transformation \(\delta\phi = i\alpha\phi\), \(\delta\phi^* = -i\alpha\phi^*\):

\[ \delta(\partial_\mu\phi) = i\alpha\,\partial_\mu\phi, \qquad \delta(\partial_\mu\phi^*) = -i\alpha\,\partial_\mu\phi^* \]

Computing the variation of \(\mathcal{L} = \partial_\mu\phi^*\,\partial^\mu\phi - m^2\phi^*\phi\):

\[ \delta\mathcal{L} = \delta(\partial_\mu\phi^*)\,\partial^\mu\phi + \partial_\mu\phi^*\,\delta(\partial^\mu\phi) - m^2[\delta(\phi^*)\,\phi + \phi^*\,\delta\phi] \]

Substituting each term:

\[ = (-i\alpha\,\partial_\mu\phi^*)\,\partial^\mu\phi + \partial_\mu\phi^*\,(i\alpha\,\partial^\mu\phi) - m^2[(-i\alpha\phi^*)\phi + \phi^*(i\alpha\phi)] \]
\[ = -i\alpha\,\partial_\mu\phi^*\,\partial^\mu\phi + i\alpha\,\partial_\mu\phi^*\,\partial^\mu\phi - m^2[-i\alpha\phi^*\phi + i\alpha\phi^*\phi] \]
\[ = 0 + 0 = 0 \]
\[ \boxed{\delta\mathcal{L} = 0} \]

(b) Derivation of the Noether Current:

For a complex scalar field, the Noether current is:

\[ j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta\phi + \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi^*)}\,\delta\phi^* \]

Computing each partial derivative. From \(\mathcal{L} = \eta^{\alpha\beta}\partial_\alpha\phi^*\,\partial_\beta\phi - m^2\phi^*\phi\):

\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)} = \partial^\mu\phi^* \]
\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi^*)} = \partial^\mu\phi \]

Substituting:

\[ j^\mu = \partial^\mu\phi^*\cdot(i\alpha\phi) + \partial^\mu\phi\cdot(-i\alpha\phi^*) \]
\[ = i\alpha\left[\phi\,\partial^\mu\phi^* - \phi^*\,\partial^\mu\phi\right] \]

Extracting the coefficient of \(\alpha\) to define the conserved current:

\[ \boxed{j^\mu = i\left(\phi\,\partial^\mu\phi^* - \phi^*\,\partial^\mu\phi\right)} \]

(c) Conserved Charge:

\[ Q = \int d^3x\,j^0 = i\int d^3x\left(\phi\,\partial^0\phi^* - \phi^*\,\partial^0\phi\right) = i\int d^3x\left(\phi\,\dot{\phi}^* - \phi^*\,\dot{\phi}\right) \]
\[ \boxed{Q = i\int d^3x\left(\phi\,\dot{\phi}^* - \phi^*\,\dot{\phi}\right)} \]

Physical meaning: After quantization, the complex scalar field describes both particles and antiparticles. When the field \(\phi\) is Fourier expanded, the positive-frequency part corresponds to particle annihilation operators, while the negative-frequency part corresponds to antiparticle creation operators. After quantization, the conserved charge \(Q\) becomes an operator proportional to "particle number \(-\) antiparticle number" (\(N_{\text{particle}} - N_{\text{antiparticle}}\)). This is a conserved quantity corresponding to \(U(1)\) symmetry, and when coupled to electromagnetic interactions, it corresponds to charge conservation.

Consistency check: If \(\phi\) is a real scalar field (\(\phi = \phi^*\)), then \(j^\mu = i(\phi\,\partial^\mu\phi - \phi\,\partial^\mu\phi) = 0\), which is consistent with the fact that a real scalar field carries no \(U(1)\) charge.

M-3. Spacetime Translational Invariance and the Energy-Momentum Tensor

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Problem:

Suppose the Lagrangian density \(\mathcal{L}(\phi, \partial_\mu\phi)\) is invariant under spacetime translations \(x^\mu \to x^\mu + a^\mu\) (where \(a^\mu\) is an infinitesimal constant vector). Under this transformation, the field changes as \(\delta\phi = -a^\nu\partial_\nu\phi\).

(a) Show that the change in the Lagrangian density itself can be written in the form of a total derivative: \(\delta\mathcal{L} = -a^\nu\partial_\nu\mathcal{L} = \partial_\mu(-a^\nu\delta^\mu{}_\nu\mathcal{L})\).

(b) Using the generalization of Noether's theorem (when \(\delta\mathcal{L} = \partial_\mu K^\mu\), the conserved current is \(j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\delta\phi - K^\mu\)), derive the canonical energy-momentum tensor

\[ T^{\mu}{}_\nu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\partial_\nu\phi - \delta^\mu{}_\nu\,\mathcal{L} \]

(c) For the Klein-Gordon field with Lagrangian density \(\mathcal{L} = \frac{1}{2}\partial_\alpha\phi\,\partial^\alpha\phi - \frac{1}{2}m^2\phi^2\), compute \(T^{00}\) and verify that it coincides with the Hamiltonian density \(\mathcal{H}\).

(a) Proof that \(\delta\mathcal{L}\) is a total derivative:

Since \(\mathcal{L}(\phi, \partial_\mu\phi)\) does not depend explicitly on \(x^\mu\), the spacetime dependence of \(\mathcal{L}\) arises only through the field \(\phi(x)\). Under a spacetime translation \(x^\mu \to x^\mu + a^\mu\), the field transforms as \(\delta\phi = -a^\nu\partial_\nu\phi\).

We compute the variation of \(\mathcal{L}\) directly. Viewing \(\mathcal{L}\) as a composite function of \(x\):

\[ \delta\mathcal{L} = \frac{\partial\mathcal{L}}{\partial\phi}\,\delta\phi + \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta(\partial_\mu\phi) \]

On the other hand, since \(\mathcal{L}\) itself is a scalar field, under the same translation:

\[ \delta\mathcal{L} = -a^\nu\partial_\nu\mathcal{L} \]

We rewrite this in the form of a total derivative. Since \(a^\nu\) is constant:

\[ \delta\mathcal{L} = -a^\nu\partial_\nu\mathcal{L} = -\partial_\mu(a^\nu\delta^\mu{}_\nu\mathcal{L}) = \partial_\mu(-a^\nu\delta^\mu{}_\nu\mathcal{L}) \]

Here we used \(\partial_\mu\delta^\mu{}_\nu = 0\) (constant) and \(\partial_\mu a^\nu = 0\) (constant).

\[ \boxed{\delta\mathcal{L} = \partial_\mu(-a^\nu\delta^\mu{}_\nu\mathcal{L})} \]

This is of the form \(\delta\mathcal{L} = \partial_\mu K^\mu\), where \(K^\mu = -a^\nu\delta^\mu{}_\nu\mathcal{L}\).

(b) Derivation of the energy-momentum tensor:

According to the generalized Noether's theorem (for the case \(\delta\mathcal{L} = \partial_\mu K^\mu\)), the conserved current is:

\[ j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta\phi - K^\mu \]

Substituting \(\delta\phi = -a^\nu\partial_\nu\phi\) and \(K^\mu = -a^\nu\delta^\mu{}_\nu\mathcal{L}\):

\[ j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,(-a^\nu\partial_\nu\phi) - (-a^\nu\delta^\mu{}_\nu\mathcal{L}) \]
\[ = -a^\nu\left[\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\partial_\nu\phi - \delta^\mu{}_\nu\mathcal{L}\right] \]

Since \(a^\nu\) is an arbitrary constant vector, we obtain an independent conserved current for each \(\nu\). Setting \(j^\mu = -a^\nu T^\mu{}_\nu\):

\[ \boxed{T^{\mu}{}_\nu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\partial_\nu\phi - \delta^\mu{}_\nu\,\mathcal{L}} \]

The conservation law is \(\partial_\mu T^{\mu}{}_\nu = 0\) (for each \(\nu\)).

(c) Computation of \(T^{00}\) for the Klein-Gordon field:

For \(\mathcal{L} = \frac{1}{2}\partial_\alpha\phi\,\partial^\alpha\phi - \frac{1}{2}m^2\phi^2\):

\[ T^{0}{}_{0} = \frac{\partial\mathcal{L}}{\partial(\partial_0\phi)}\,\partial_0\phi - \delta^0{}_{0}\,\mathcal{L} \]

Since \(\frac{\partial\mathcal{L}}{\partial(\partial_0\phi)} = \partial^0\phi = \dot{\phi}\):

\[ T^{0}{}_{0} = \dot{\phi}\cdot\dot{\phi} - \mathcal{L} = \dot{\phi}^2 - \left[\frac{1}{2}\dot{\phi}^2 - \frac{1}{2}(\nabla\phi)^2 - \frac{1}{2}m^2\phi^2\right] \]
\[ = \dot{\phi}^2 - \frac{1}{2}\dot{\phi}^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{1}{2}m^2\phi^2 \]
\[ T^{00} = \frac{1}{2}\dot{\phi}^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{1}{2}m^2\phi^2 \]

Comparing with the Hamiltonian density obtained in S1(b) (using \(\pi = \dot{\phi}\)):

\[ \mathcal{H} = \frac{1}{2}\pi^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{1}{2}m^2\phi^2 = \frac{1}{2}\dot{\phi}^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{1}{2}m^2\phi^2 \]
\[ \boxed{T^{00} = \mathcal{H}} \]

Consistency check: \(T^{00}\) is the energy density and should coincide with the Hamiltonian density. Moreover, \(T^{00} \geq 0\) (each term is positive definite), confirming that the energy is bounded from below. Furthermore, computing \(T^{0i}\) gives \(T^{0i} = \dot{\phi}\,\partial_i\phi\), which corresponds to the momentum density.

M-4. Euler-Lagrange Equations for the Dirac Field

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Problem:

The Lagrangian density for the Dirac field \(\psi(x)\) (a 4-component spinor) is given by

\[ \mathcal{L}_{\mathrm{Dirac}} = \bar{\psi}(i\gamma^\mu\partial_\mu - m)\psi \]

where \(\bar{\psi} = \psi^\dagger\gamma^0\) is the Dirac conjugate and \(\gamma^\mu\) are the gamma matrices.

(a) Treating \(\bar{\psi}\) as an independent variable, derive the Euler-Lagrange equation with respect to \(\bar{\psi}\) and obtain the Dirac equation

\[ (i\gamma^\mu\partial_\mu - m)\psi = 0 \]

(b) Derive the Euler-Lagrange equation with respect to \(\psi\) and obtain the adjoint Dirac equation for \(\bar{\psi}\)

\[ \bar{\psi}(i\overleftarrow{\partial}_\mu\gamma^\mu + m) = 0 \]

where \(\overleftarrow{\partial}_\mu\) denotes a derivative acting to the left.

(a) Euler-Lagrange Equation with Respect to \(\bar{\psi}\):

\[ \mathcal{L}_{\mathrm{Dirac}} = \bar{\psi}(i\gamma^\mu\partial_\mu - m)\psi = i\bar{\psi}\gamma^\mu\partial_\mu\psi - m\bar{\psi}\psi \]

We treat \(\bar{\psi}\) as an independent variable. There is no term containing \(\partial_\mu\bar{\psi}\) in \(\mathcal{L}\) (the \(\partial_\mu\) acts only on \(\psi\)). Therefore:

\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\bar{\psi})} = 0 \]

The Euler-Lagrange equation is:

\[ \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\bar{\psi})}\right) - \frac{\partial\mathcal{L}}{\partial\bar{\psi}} = 0 \implies -\frac{\partial\mathcal{L}}{\partial\bar{\psi}} = 0 \]

Computing the partial derivative with respect to \(\bar{\psi}\):

\[ \frac{\partial\mathcal{L}}{\partial\bar{\psi}} = i\gamma^\mu\partial_\mu\psi - m\psi = (i\gamma^\mu\partial_\mu - m)\psi \]

Therefore:

\[ \boxed{(i\gamma^\mu\partial_\mu - m)\psi = 0} \]

This is the Dirac equation.

(b) Euler-Lagrange Equation with Respect to \(\psi\):

We treat \(\psi\) as an independent variable.

Partial derivative with respect to \(\psi\):

\[ \frac{\partial\mathcal{L}}{\partial\psi} = -m\bar{\psi} \]

(The term \(i\bar{\psi}\gamma^\mu\partial_\mu\psi\) contains \(\partial_\mu\psi\), so it does not contribute to the partial derivative with respect to \(\psi\) itself.)

Partial derivative with respect to \(\partial_\mu\psi\):

\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\psi)} = i\bar{\psi}\gamma^\mu \]

Substituting into the Euler-Lagrange equation:

\[ \partial_\mu\left(i\bar{\psi}\gamma^\mu\right) - (-m\bar{\psi}) = 0 \]
\[ i(\partial_\mu\bar{\psi})\gamma^\mu + m\bar{\psi} = 0 \]

Rewriting using the left-acting derivative notation \(\overleftarrow{\partial}_\mu\):

\[ \boxed{\bar{\psi}(i\overleftarrow{\partial}_\mu\gamma^\mu + m) = 0} \]

This is the adjoint Dirac equation.

Verification: We take the Dirac conjugate of the Dirac equation \((i\gamma^\mu\partial_\mu - m)\psi = 0\). Taking the Hermitian conjugate and multiplying by \(\gamma^0\) from the right:

\[ [(i\gamma^\mu\partial_\mu - m)\psi]^\dagger\gamma^0 = 0 \]
\[ \psi^\dagger(-i\gamma^{\mu\dagger}\overleftarrow{\partial}_\mu - m)\gamma^0 = 0 \]

Using the relation \(\gamma^0\gamma^{\mu\dagger}\gamma^0 = \gamma^\mu\) (the Hermiticity property of gamma matrices), we obtain \(\bar{\psi}(i\overleftarrow{\partial}_\mu\gamma^\mu + m) = 0\), which is consistent with the result of (b).

Advanced

A-1. Generalization of Noether's Theorem

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Problem:

Consider the case where, under the field transformation \(\phi \to \phi + \delta\phi\), the Lagrangian density is not invariant but changes by a total derivative (4-dimensional divergence) expressed using some 4-vector \(K^\mu\):

\[ \delta\mathcal{L} = \partial_\mu K^\mu \]

(a) Show that the variation \(\delta S\) of the action \(S = \int d^4x\,\mathcal{L}\) vanishes in this case as well (under appropriate boundary conditions), and confirm that the equations of motion remain unchanged.

(b) Prove that the modified Noether current

\[ j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta\phi - K^\mu \]

satisfies \(\partial_\mu j^\mu = 0\) when the equations of motion hold.

(c) As an application, consider a Lorentz boost transformation (in the \(x\) direction with infinitesimal rapidity \(\delta\omega\)) for the real scalar field \(\mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi - V(\phi)\):

\[ \delta\phi = -\delta\omega\,(t\,\partial_x\phi + x\,\partial_t\phi) \]

Construct the corresponding conserved current. Discuss the physical meaning of the conserved charge.

(a) Invariance of the Action:

The variation of the action is:

\[ \delta S = \int d^4x\,\delta\mathcal{L} = \int d^4x\,\partial_\mu K^\mu \]

Applying the 4-dimensional Gauss's theorem (divergence theorem):

\[ \delta S = \int d^4x\,\partial_\mu K^\mu = \oint_{\partial\Omega} K^\mu\,dS_\mu \]

where \(\partial\Omega\) is the boundary surface of the 4-dimensional integration region. Under the boundary condition that the field variation \(\delta\phi\) vanishes at the boundary (or that \(K^\mu\) decays sufficiently rapidly at the boundary):

\[ \boxed{\delta S = 0} \]

Therefore, the equations of motion (the Euler-Lagrange equations derived from \(\delta S = 0\)) remain unchanged.

(b) Proof of Conservation of the Modified Noether Current:

Computing \(\delta\mathcal{L}\) through the field variation:

\[ \delta\mathcal{L} = \frac{\partial\mathcal{L}}{\partial\phi}\,\delta\phi + \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\partial_\mu(\delta\phi) \]

Rewriting the second term via integration by parts:

\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\partial_\mu(\delta\phi) = \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta\phi\right) - \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right)\delta\phi \]

Substituting:

\[ \delta\mathcal{L} = \left[\frac{\partial\mathcal{L}}{\partial\phi} - \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right)\right]\delta\phi + \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta\phi\right) \]

When the equations of motion hold, the term in square brackets vanishes:

\[ \delta\mathcal{L} = \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta\phi\right) \]

On the other hand, by assumption \(\delta\mathcal{L} = \partial_\mu K^\mu\). Equating both sides:

\[ \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta\phi\right) = \partial_\mu K^\mu \]
\[ \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta\phi - K^\mu\right) = 0 \]
\[ \boxed{\partial_\mu j^\mu = 0, \qquad j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta\phi - K^\mu} \]

(c) Application to Lorentz Boost Transformations:

Under an infinitesimal Lorentz boost in the \(x\) direction (rapidity \(\delta\omega\)):

\[ \delta x^0 = -\delta\omega\,x^1, \qquad \delta x^1 = -\delta\omega\,x^0, \qquad \delta x^2 = \delta x^3 = 0 \]

From the scalar field transformation rule \(\delta\phi = -\delta x^\nu\partial_\nu\phi\):

\[ \delta\phi = -(-\delta\omega\,x^1)\partial_0\phi - (-\delta\omega\,x^0)\partial_1\phi = \delta\omega(x^1\partial_0\phi + x^0\partial_1\phi) \]

That is, \(\delta\phi = \delta\omega(x\,\dot{\phi} + t\,\partial_x\phi)\). (This is consistent with the sign convention in the problem statement \(\delta\phi = -\delta\omega(t\,\partial_x\phi + x\,\partial_t\phi)\), where \(x^0 = t\), \(x^1 = x\).)

Determination of \(K^\mu\):

Since \(\mathcal{L}\) is also a scalar, under the same transformation:

\[ \delta\mathcal{L} = -\delta x^\nu\partial_\nu\mathcal{L} = \delta\omega(x^1\partial_0\mathcal{L} + x^0\partial_1\mathcal{L}) \]

Writing this in total derivative form:

\[ \delta\mathcal{L} = \delta\omega\left[\partial_0(x^1\mathcal{L}) + \partial_1(x^0\mathcal{L})\right] - \delta\omega\left[(\partial_0 x^1)\mathcal{L} + (\partial_1 x^0)\mathcal{L}\right] \]

Since \(\partial_0 x^1 = 0\) and \(\partial_1 x^0 = 0\):

\[ \delta\mathcal{L} = \delta\omega\,\partial_\mu\left[\delta^\mu{}_0\,x^1\mathcal{L} + \delta^\mu{}_1\,x^0\mathcal{L}\right] = \partial_\mu K^\mu \]

where \(K^\mu = \delta\omega(\delta^\mu{}_0\,x^1\mathcal{L} + \delta^\mu{}_1\,x^0\mathcal{L})\), i.e., \(K^0 = \delta\omega\,x^1\mathcal{L}\), \(K^1 = \delta\omega\,x^0\mathcal{L}\), \(K^2 = K^3 = 0\).

Construction of the conserved current:

\[ j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta\phi - K^\mu \]

Using \(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)} = \partial^\mu\phi\):

\[ j^\mu = \partial^\mu\phi\cdot\delta\omega(x^1\partial_0\phi + x^0\partial_1\phi) - \delta\omega(\delta^\mu{}_0\,x^1\mathcal{L} + \delta^\mu{}_1\,x^0\mathcal{L}) \]

Factoring out \(\delta\omega\):

\[ j^\mu = \partial^\mu\phi(x^1\partial_0\phi + x^0\partial_1\phi) - \delta^\mu{}_0\,x^1\mathcal{L} - \delta^\mu{}_1\,x^0\mathcal{L} \]

Rewriting in terms of the energy-momentum tensor \(T^{\mu}{}_\nu = \partial^\mu\phi\,\partial_\nu\phi - \delta^\mu{}_\nu\mathcal{L}\):

\[ j^\mu = x^1(\partial^\mu\phi\,\partial_0\phi - \delta^\mu{}_0\mathcal{L}) + x^0(\partial^\mu\phi\,\partial_1\phi - \delta^\mu{}_1\mathcal{L}) \]
\[ = x^1 T^{\mu 0} + x^0 T^{\mu 1} \]

Here we used \(T^{\mu\nu} = \eta^{\nu\alpha}T^\mu{}_\alpha\). Noting that \(T^{\mu 0} = T^\mu{}_0\) (since \(\eta^{00} = 1\)) and \(T^{\mu 1} = -T^\mu{}_1\) (since \(\eta^{11} = -1\)):

\[ j^\mu = x^1 T^{\mu 0} - x^0 T^{\mu 1} \]

Alternatively, as the conserved current corresponding to the generator of the Lorentz transformation:

\[ \boxed{M^{\mu 01} = x^0 T^{\mu 1} - x^1 T^{\mu 0}} \]

(The sign convention follows \(M^{\mu\rho\sigma} = x^\rho T^{\mu\sigma} - x^\sigma T^{\mu\rho}\).)

Conserved charge:

\[ M^{01} = \int d^3x\,M^{001} = \int d^3x\,(x^0 T^{01} - x^1 T^{00}) = t\,P^1 - \int d^3x\,x\,\mathcal{H} \]

where \(P^1 = \int d^3x\,T^{01}\) is the total momentum in the \(x\) direction.

Physical meaning: This conserved charge is the boost generator and contains information about the "center of energy" of the system. The conservation of \(M^{01} = tP^1 - \int d^3x\,x\,\mathcal{H}\) means:

\[ \frac{d}{dt}\left(tP^1 - \int d^3x\,x\,\mathcal{H}\right) = 0 \]
\[ P^1 + t\frac{dP^1}{dt} - \frac{d}{dt}\int d^3x\,x\,\mathcal{H} = 0 \]

Using the conservation of \(P^1\) (\(\frac{dP^1}{dt} = 0\)), we obtain \(\frac{d}{dt}\int d^3x\,x\,\mathcal{H} = P^1\). This means that "the center of energy moves at constant velocity," which is the field theory version of the fact that in particle mechanics, \(X_{\mathrm{cm}} = \frac{\sum m_i x_i}{\sum m_i}\) undergoes uniform rectilinear motion.

Verification: We directly confirm \(\partial_\mu M^{\mu\rho\sigma} = 0\). \(\partial_\mu M^{\mu\rho\sigma} = \partial_\mu(x^\rho T^{\mu\sigma} - x^\sigma T^{\mu\rho}) = \delta^\rho_\mu T^{\mu\sigma} + x^\rho\partial_\mu T^{\mu\sigma} - \delta^\sigma_\mu T^{\mu\rho} - x^\sigma\partial_\mu T^{\mu\rho} = T^{\rho\sigma} - T^{\sigma\rho}\) (using \(\partial_\mu T^{\mu\nu} = 0\)). This is conserved if the canonical energy-momentum tensor is symmetric, \(T^{\rho\sigma} = T^{\sigma\rho}\). For the Klein-Gordon field, \(T^{\mu\nu} = \partial^\mu\phi\,\partial^\nu\phi - \eta^{\mu\nu}\mathcal{L}\) is indeed symmetric, which is consistent.

A-2. Lagrangian of the Maxwell Field and Gauge Invariance

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Problem:

The Lagrangian density of the electromagnetic field is given by

\[ \mathcal{L}_{\mathrm{Maxwell}} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} \]

where the field strength tensor is \(F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu\), and \(A_\mu(x)\) is the 4-vector potential.

(a) Derive the Euler-Lagrange equations for \(A_\nu\) and obtain the vacuum Maxwell equations (without sources)

\[ \partial_\mu F^{\mu\nu} = 0 \]

Explicitly show the computation of \(\dfrac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)}\) in the derivation.

(b) Show that \(F_{\mu\nu}\) is invariant under the gauge transformation \(A_\mu(x) \to A_\mu(x) + \partial_\mu\Lambda(x)\) (where \(\Lambda(x)\) is an arbitrary scalar function), and thereby confirm that \(\mathcal{L}_{\mathrm{Maxwell}}\) is also gauge invariant.

(c) Discuss what happens when one attempts to directly apply Noether's theorem to the gauge transformation. In particular, explain how the fact that the gauge transformation parameter \(\Lambda(x)\) is a function of spacetime (a local transformation) relates to Noether's theorem, which assumes a global symmetry, and mention the existence of Noether's second theorem.

(a) Derivation of the Euler-Lagrange Equations:

\(\mathcal{L}_{\mathrm{Maxwell}} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\) with \(F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu\).

Step 1: Computing \(\frac{\partial\mathcal{L}}{\partial A_\nu}\)

Since \(\mathcal{L}\) does not contain \(A_\nu\) itself (without derivatives):

\[ \frac{\partial\mathcal{L}}{\partial A_\nu} = 0 \]

Step 2: Computing \(\frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)}\)

Expanding \(F_{\mu\nu}F^{\mu\nu}\):

\[ F_{\alpha\beta}F^{\alpha\beta} = F_{\alpha\beta}\,\eta^{\alpha\gamma}\eta^{\beta\delta}F_{\gamma\delta} = (\partial_\alpha A_\beta - \partial_\beta A_\alpha)\eta^{\alpha\gamma}\eta^{\beta\delta}(\partial_\gamma A_\delta - \partial_\delta A_\gamma) \]

Computing the partial derivative with respect to \(\partial_\mu A_\nu\). Since \(F_{\alpha\beta} = \partial_\alpha A_\beta - \partial_\beta A_\alpha\):

\[ \frac{\partial F_{\alpha\beta}}{\partial(\partial_\mu A_\nu)} = \delta^\mu_\alpha\delta^\nu_\beta - \delta^\mu_\beta\delta^\nu_\alpha \]

Using the product rule:

\[ \frac{\partial}{\partial(\partial_\mu A_\nu)}(F_{\alpha\beta}F^{\alpha\beta}) = \frac{\partial F_{\alpha\beta}}{\partial(\partial_\mu A_\nu)}F^{\alpha\beta} + F_{\alpha\beta}\frac{\partial F^{\alpha\beta}}{\partial(\partial_\mu A_\nu)} \]

Since \(F^{\alpha\beta} = \eta^{\alpha\gamma}\eta^{\beta\delta}F_{\gamma\delta}\), the second term has the same structure. By symmetry, both terms are equal:

\[ \frac{\partial}{\partial(\partial_\mu A_\nu)}(F_{\alpha\beta}F^{\alpha\beta}) = 2(\delta^\mu_\alpha\delta^\nu_\beta - \delta^\mu_\beta\delta^\nu_\alpha)F^{\alpha\beta} = 2(F^{\mu\nu} - F^{\nu\mu}) = 4F^{\mu\nu} \]

where we used \(F^{\mu\nu} = -F^{\nu\mu}\) (antisymmetry).

Therefore:

\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)} = -\frac{1}{4}\cdot 4F^{\mu\nu} = -F^{\mu\nu} \]

Step 3: The Euler-Lagrange Equations

\[ \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)}\right) - \frac{\partial\mathcal{L}}{\partial A_\nu} = 0 \]
\[ \partial_\mu(-F^{\mu\nu}) - 0 = 0 \]
\[ \boxed{\partial_\mu F^{\mu\nu} = 0} \]

These are the Maxwell equations in vacuum (source-free). Written in components, \(\nu = 0\) corresponds to \(\nabla\cdot\mathbf{E} = 0\) (Gauss's law), and \(\nu = i\) corresponds to \(-\frac{\partial\mathbf{E}}{\partial t} + \nabla\times\mathbf{B} = 0\) (Ampère's law).

(b) Proof of Gauge Invariance:

Computing the change in the field strength tensor under the gauge transformation \(A_\mu \to A'_\mu = A_\mu + \partial_\mu\Lambda\):

\[ F'_{\mu\nu} = \partial_\mu A'_\nu - \partial_\nu A'_\mu = \partial_\mu(A_\nu + \partial_\nu\Lambda) - \partial_\nu(A_\mu + \partial_\mu\Lambda) \]
\[ = \partial_\mu A_\nu + \partial_\mu\partial_\nu\Lambda - \partial_\nu A_\mu - \partial_\nu\partial_\mu\Lambda \]

By the commutativity of partial derivatives \(\partial_\mu\partial_\nu\Lambda = \partial_\nu\partial_\mu\Lambda\):

\[ F'_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu = F_{\mu\nu} \]
\[ \boxed{F'_{\mu\nu} = F_{\mu\nu}} \]

Since \(F_{\mu\nu}\) is gauge invariant, \(\mathcal{L}_{\mathrm{Maxwell}} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\) is also gauge invariant.

(c) Relationship Between Noether's Theorem and Gauge Transformations:

Analysis of the Issue:

Noether's first theorem applies to global continuous symmetries, i.e., transformations whose parameters are constants that do not depend on spacetime. However, the parameter \(\Lambda(x)\) of the gauge transformation \(A_\mu \to A_\mu + \partial_\mu\Lambda(x)\) is an arbitrary function of spacetime (a local parameter).

If we attempt to directly apply Noether's theorem:

  1. Global gauge transformation (\(\Lambda = \text{const}\)): In this case \(\delta A_\mu = \partial_\mu\Lambda = 0\), so the field does not change. Therefore, only the trivial identity \(0 = 0\) is obtained, and no non-trivial conserved current can be constructed.

  2. Local gauge transformation (\(\Lambda(x)\) is an arbitrary function): Since the parameter has infinitely many degrees of freedom, it cannot be handled within the framework of the ordinary Noether theorem.

Noether's Second Theorem:

When the transformation parameters are arbitrary functions of spacetime (local symmetry), Noether's second theorem applies. This theorem states:

If the action is invariant under transformations involving an arbitrary function \(\Lambda(x)\) and its finite-order derivatives, then there exist identities (analogous to Bianchi identities) among the Euler-Lagrange equations.

In the case of Maxwell theory, this identity is:

\[ \partial_\nu(\partial_\mu F^{\mu\nu}) \equiv 0 \]

which holds identically due to the antisymmetry of \(F^{\mu\nu}\), without using the equations of motion.

Physical Meaning:

  • Gauge symmetry is not a physical symmetry (one that generates new conserved quantities), but rather reflects a redundancy in the description of the theory.
  • Of the 4 components of \(A_\mu\), only 2 are physical degrees of freedom (the two polarization states of the photon), and the remaining 2 correspond to gauge redundancy.
  • While Noether's first theorem gives "symmetry → conserved quantity," Noether's second theorem gives "local symmetry → identities among equations of motion (constraints)."
\[ \boxed{\text{Gauge symmetry (local symmetry)} \xrightarrow{\text{Noether's 2nd theorem}} \text{Identities among equations of motion + reduction of physical degrees of freedom}} \]

Verification: Counting the physical degrees of freedom in Maxwell theory. \(A_\mu\) has 4 components, but due to gauge freedom (one arbitrary function \(\Lambda(x)\)), 1 component is redundant. Furthermore, \(\partial_\mu F^{\mu 0} = 0\) provides a constraint (a condition on initial values), removing another component. The result is \(4 - 1 - 1 = 2\) degrees of freedom, corresponding to the two physical polarizations (transverse waves) of the photon.